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2nd half of class starting with Inferences for two population means using paired samples

State one possible advantage of using paired samples instead of independent samples.

By using a paired sample, extraneous sources of variation can be removed. The sampling error thus made in estimating the difference between the population means will generally be smaller. As a result, detecting differences between the population means is more likely when such differences exist.

State the two conditions required for performing a paired t-procedure. How important are these conditions?

(1) simple random paired sample
(2) normal differences or large sample

The simple random paired sample is essential. Moderate violations of the normal differences assumption are permissible even for small or moderate size samples.

The U.S. Census Bureau publishes information on the ages of married people. Suppose you want to use a paired sample to decide whether the mean age of married men exceeds the mean age of married women.
(a) Identify the variable
(b) Identify the two populations
(c) Identify the pairs
(d) Identify the paired-difference variable
(e) Determine the null and alternative hypotheses
(f) Classify the hypothesis test as two tailed, left tailed, or right tailed.

(a) age
(b) married men and married women
(c) married couples
(d) the difference between the ages of a married couple
(e) H₀: µ₁ = µ₂
H₁: µ₁ > µ₂
(f) right tailed

How to calculate test statistic for paired t-test

df = n-1

How to calculate critical value(s) for paired t-test

df = n-1

C. Tu examined the effects of construction of new sports stadiums on home values. Suppose that you want to use a paired sample to decide whether construction of new sports stadiums affects the mean price of neighboring homes.
(a) Identify the variable
(b) Identify the two populations
(c) Identify the pairs
(d) Identify the paired-difference variable
(e) Determine the null and alternative hypotheses
(f) Classify the hypothesis test as two tailed, left tailed, or right tailed.

(a) home price
(b) homes neighboring and homes not neighboring newly constructed sports stadiums
(c) a pair of comparable homes, one neighboring and the other not neighboring a newly constructed sports stadium
(d) the difference between the prices of a pair of comparable homes, one neighboring and the other not neighboring a newly constructed sports stadium
(e) H₀: µ₁ = µ₂
H₁: µ₁ ≠ µ₂
where µ₁ = mean price of homes neighboring
and where µ₂ = mean price of homes not neighboring
(f) two tailed

Charles Darwin investigated the effect of cross-fertilization on the heights of plants. In one study he planted 15 pairs of Zea mays plants. Each pair consisted of one cross-fertilized plant and one self-fertilized plant grown in the same pot. The following table gives the height differences, in eighths of an inch, for the 15 pairs. Each difference is obtained by subtracting the height of the self-fertilized plant from that of the cross-fertilized plant (see table in Exercise 10.119).
(a) Identify the variable under consideration.
(b) Identify the two populations.
(c) Identify the paired-difference variable
(d) Are the numbers in the table paired differences? Why or why not?
(e) At the 5% significance level, do the data provide sufficient evidence to conclude that the mean heights of cross-fertilized and self-fertilized Zea mays differ? (Note: d-bar = 20.93 and Sd = 37.74).

(a) height (of Zea mays)
(b) Cross-fertilized Zea mays and self-fertilized Zea mays
(c) The difference between the heights of a cross-fertilized Zea may and a self-fertilized Zea may grown in the same pot
(d) Yes. Because each number is the difference between the heights of a cross-fertilized Zea may and a self-fertilized Zea may grown in the same pot.
(e) H₀: µ₁ = µ₂, H₁: µ₁ ≠ µ₂
∝ = 0.05
t = 2.148
critical values: = ± 2.145
reject H₀
At the 5% significance level, the data provide sufficient evidence to conclude that the mean heights of cross-fertilized and self-fertilized Zea mays differ.

Anorexia nervosa is a serious eating disorder, particularly among young women. The following data provide the weights, in pounds, of 17 anorexic young women before and after receiving a family therapy treatment for anorexia nervosa (see 10.121, p. 546 for data). Does family therapy appear to be effective in helping anorexic young women gain weight? Perform the appropriate hypothesis test at the 5% significance level.

H₀: µ₁ = µ₂, H₁: µ₁ < µ₂
∝ = 0.05
t = -4.185
critical value: = -1.746
Reject H₀
At the 5% significance level, the data provide sufficient evidence to conclude that family therapy is effective in helping anorexic young women gain weight.

Stichler, Richey, and Mandel compared two methods of measuring tire treadwear. Eleven tires were each measured for treadwear by two methods, one based on weight and the other on groove wear. The following are the data, in thousands of miles (see 10.122, P. 546 for data).
At the 5% significance level, do the data provide sufficient evidence to conclude that, on average, the two measurement methods give different results?

(1) H₀: µ₁ = µ₂
H₁: µ₁ ≠ µ₂
(2) ∝ = 0.05
(3) test statistic: 3.87; p-value = .0031
(4) Reject H₀
(5) There is sufficient evidence to conclude that, on average, the two measurement methods give different results.

Formula for confidence interval for paired differences (paired t-interval)

In this problem, d-bar is positive, meaning that the weight method overestimates or the groove method underestimates.

In a paired-difference confidence interval, what does it mean when 0 (zero) does not lie within the interval?

It means you can reject H₀. This is another hypothesis test method by using confidence interval. If 0 is in the interval, do not reject H₀.

Glaucoma is a leading cause of blindness in the U.S. N. Ehlers measured the corneal thickness of 8 patients who had glaucoma in one eye but not in the other. The following are the data on corneal thickness , in microns (see 10.123 P. 546 for data). At the 10% significance level, do the data provide sufficient evidence to conclude that mean corneal thickness is greater in normal eyes than in eyes with glaucoma?

(1) H₀: µ₁ = µ₂
H₁: µ₁ > µ₂
(2) ∝ = 0.10
(3) test statistic: 1.053
(4) critical value = 1.415
(4) Do not reject H₀
(5) There is sufficient evidence to conclude that, on average, the two measurement methods give different results.At the 10% significance level, the data do not provide sufficient evidence to conclude that mean corneal thickness is greater in normal eyes than in eyes with glaucoma.

Anorexia nervosa is a serious eating disorder, particularly among young women. The following data provide the weights, in pounds, of 17 anorexic young women before and after receiving a family therapy treatment for anorexia nervosa (see 10.121, p. 546 for data). Find a 90% confidence interval for the weight gain that would be obtained, on average, by using the family therapy treatment.

-10.30 to - 4.23 lb.
We can be 90% confident that the weight gain that would be obtained, on average, by using the family therapy treatment is between 4.23 and 10.30 lbs.

Glaucoma is a leading cause of blindness in the U.S. N. Ehlers measured the corneal thickness of 8 patients who had glaucoma in one eye but not in the other. The following are the data on corneal thickness , in microns (see 10.123 P. 546 for data). Obtain an 80% confidence interval for the difference between the mean corneal thickness of normal eyes and that of eyes with glaucoma.

-1.4 to 9.4 microns. We can be 80% confident that the difference between the mean corneal thickness of normal eyes and that of eyes with glaucoma is between -1.4 and 9.4 microns.

To assess the effects of two different strains of the tobacco mosaic virus, W. Youden and H. Beale randomly selected 8 tobacco leaves. Half of each leaf was subjected to one of the strains of tobacco mosaic virus and the other half to the other strain. The researchers then counted the number of local lesions apparent on each half of each leaf. The data are as follows: (see 10.131 P. 547 for data). Suppose that you want to perform a hypothesis test to determine whether a difference exists between the mean numbers of local lesions resulting from the two viral strains. Conduct preliminary graphical analyses to decide whether applying the paired t-test is reasonable. Explain your decision.

(1) look at the sample size. If it is small (less than 30), we need to examine issues of normality and outliers.
(2) construct a normal probability plot of the paired differences
Are there outliers?
Is the plot roughly linear?
In this problem, there is one outlier, but is otherwise roughly linear.
Therefore, in view of the small sample size, applying the paired t-test is not reasonable (because of the outlier).

Suppose that you want to perform a hypothesis test based on independent simple random samples to compare the means of two populations. Assume that the variable under consideration is normally distributed on each of the two populations and that the population standard deviations are equal. Which hypothesis test procedure is the best one to use?

the pooled t-test

Suppose that you want to perform a hypothesis test based on independent simple random samples to compare the means of two populations. Assume that the variable under consideration is normally distributed on each of the two populations and that the population standard deviations are not equal. Which hypothesis test procedure is the best one to use?

the nonpooled t-test

Suppose that you want to perform a hypothesis test based on a simple random paired sample to compare the means of two populations. Assume that the paired-difference variable is normally distributed. Which hypothesis test procedure is the best one to use?

the paired t-test

What are the two primary hypothesis testing procedures based on independent simple random samples to compare the means of two populations with unknown standard deviations?

the pooled t-test and the nonpooled t-test

List the conditions for using the pooled t-test

1. simple random sample
2. independent samples
3. normal populations or large samples
4. equal population standard deviations

List the conditions for using the nonpooled t-test

1. simple random sample
2. independent samples
3. normal populations or large samples
4. unequal population standard deviations

For both pooled and nonpooled t-tests, the population standard deviations are unknown. How do you decide which test to use?

Compare the sample standard deviations. If they are equal or close, choose the pooled t-test. If one is twice the other, use the nonpooled t-test.

J. Misner published results of a study on grip and leg strength of males and females. The following data, in newtons, is based on their measurements of right-leg strength (see Problem 7, P. 567 for data). Preliminary data analyses indicate that you can reasonably presume leg strength is normally distributed for both males and females and that the standard deviations of leg strength are approximately equal.
(a) At the 5% significance level, do the data provide sufficient evidence to conclude that mean right-leg strength of males exceeds that of females? (Note: x-bar₁ = 2127, s₁ = 513, x-bar₂ = 1843, s₂ = 446.)
(b) Determine a 90% confidence interval for the difference between the mean right-leg strengths of males and females. Interpret your result.

(a) H₀: µ₁ = µ₂; H₁: µ₁ > µ₂
∝ = 0.05
t = 1.538; Sp = 479.33
Critical value: df = 25; CV = 1.708
Do not reject H₀
At the 5% significance level, the data do not provide enough evidence to conclude that the mean right-leg strength of males exceeds that of females.

(b) (-31.33, 599.33)
We can be 90% confident that the difference, µ₁ - µ₂, between the mean right-leg strengths of males and females is between -31.3 and 599.3 newtons.

Researchers examined the reproductive characteristics of the eastern cottonmouth. The data in the following table, based on the results of the researchers' study, give the number of young per litter for 24 female cottonmouths in Florida and 44 female cottonmouths in Virginia (see Problem 9, P. 567 for data). Preliminary data analyses indicate that you can reasonably presume that litter sizes of cottonmouths in both states are approximately normally distributed.
(a) At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, the number of young per litter of cottonmouths in Florida is less than that in Virginia? Do not assume that the population standard deviations are equal. (Note: x-bar₁ = 5.46, s₁ = 1.59, x-bar₂ = 7.59, s₂ = 2.68).
(b) Find a 98% confidence interval for the difference between the mean litter sizes of cottonmouths in Florida and Virginia. Interpret your result.

(a) H₀: µ₁ = µ₂; H₁: µ₁ < µ₂
∝ = 0.01
t = -4.119
Critical value: -2.385 (df = 65)
Reject H₀
At the 1% significance level, the data provides enough evidence to conclude that the average litter size of cottonmouths in Florida is less than that in Virginia

(b) (-3.37, -0.89)
We can be 98% confident that the average difference, µ₁ - µ₂, between cottonmouth litter sizes in Florida and Virginia is between -3.37 and -0.89.

R. Bohanan questioned whether there is evidence for global warming in long term data on changes in dates of ice cover in three Wisconsin Lakes. The following table gives data, for a sample of 8 years, on the number of days that ice stayed on two lakes in Madison, WI - Lake Mendota and Lake Monona (see Problem 12, P. 568 for data).
(a) At the 10% significance level, do the data provide sufficient evidence to conclude that a difference exists in the mean length of time that ice stays on these two lakes?
(b) Find a 90% confidence interval for the difference in the mean lengths of time that ice stays on the two lakes. Interpret your result.

(a) H₀: µ₁ = µ₂; H₁: µ₁ ≠ µ₂
∝ = 0.10
t = 0.549 (d-bar = 1.375, Sd = 7.090)
Critical value: ± 1.860
Do not reject H₀
At the 10% significance level, the data do not provide enough evidence to conclude that there is a difference in the mean length of time that ice stays on the two lakes.

(b) (-3.374, 6.124)
We can be 90% confident that the average difference, µ₁ - µ₂, between the mean number of days ice stays on Lakes Mendota and Monona is between -3.374 and 6.124 days.

Basic properties of χ²-curves

(1) The total area under a χ²-curve equals 1
(2) A χ²-curve starts at 0 on the horizontal axis and extends indefinitely to the right, approaching, but never touching, the horizontal axis as it does so.
(3) A χ²-curve is right skewed
(4) As the number of degrees of freedom becomes larger, χ²-curves look increasingly like normal curves

Two χ²-curves have degrees of freedom 12 and 20 respectively. Which curve more closely resembles a normal curve? Explain your answer.

The χ²-curve with degree of freedom = 20 more closely resembles a normal curve. As the number of degrees of freedom becomes larger, χ²-curves look increasingly like normal curves.

What is meant by saying that a variable has a chi-square distribution?

A variable is said to have a chi-square distribution if its distribution has the shape of a special type of right-skewed curve, called a chi-square curve.

For a χ²-curve with 19 degrees of freedom, find the χ²-value that has:
(a) 0.025 to its right
(b) 0.95 to its right

(a) 32.852
(b) 10.117

For a χ²-curve with df = 10, determine:
(a) χ² (sub 0.05)
(b) χ² (sub 0.975)

(a) 18.307
(b) 3.247

Consider a χ²-curve with df = 8. Obtain the χ²-value that has area:
(a) 0.01 to its left
(b) 0.95 to its left

(a) 1.646
(b) 15.507

Determine the two χ² values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas for a χ²-curve with
(a) df = 5
(b) df = 26

(a) 0.831, 12.833
(b) 13.844, 41.923

When you use chi-square procedures to make inferences about a population standard deviation, why should the variable under consideration be normally distributed or nearly so?

Because the procedures are based on the assumption that the variable under consideration is normally distributed and are non-robust to violations of that assumption.

Use the one-standard-deviation χ² test and the one-standard-deviation χ²-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval:
s=3 and n=10
(a) H₀: ϑ = 4; H₁: ϑ < 4, ∝ = 0.05
(b) 90% confidence interval

(a) χ² = 5.062; critical value = 3.325; P = 0.171; do not reject H₀
(b) 2.19 to 4.94

Use the one-standard-deviation χ² test and the one-standard-deviation χ²-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval:
s=7 and n=26
(a) H₀: ϑ = 5; H₁: ϑ > 5, ∝ = 0.01
(b) 98% confidence interval

(a) χ² = 49; critical value = 44.314; P = 0.003; Reject H₀
(b) 5.26 to 10.31

Use the one-standard-deviation χ² test and the one-standard-deviation χ²-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval:
s=5 and n=20
(a) H₀: ϑ = 6; H₁: ϑ ≠ 6, ∝ = 0.05
(b) 95% confidence interval

(a) χ² = 13.194; critical values = 8.907 and 32.852; P = 0.343; do not reject H₀
(b) 3.80 to 7.30

In 2000, the mean retail price of agriculture books was $66.52. This year's retail prices for 28 randomly selected agriculture books are shown in the following table: (see Exercise 11.21, P. 586 for data). Assume that the population standard deviation of prices for this year's agriculture books is $8.45. At the 10% significance level, do the data provide evidence against that assumption? (Note: s = 9.229)

(1) H₀: ϑ = $8.45; H₁ ≠ $8.45
(2) ∝ = 0.10
(3) χ² = 32.207
(4) Critical values: 16.151 and 40.113; P = 0.449
(5) Do not reject H₀
(6) At the 10% significance level, the data do not provide sufficient evidence to conclude that the population standard deviation of prices for this year's agriculture books differs from $8.45.

R. Morris and E. Watson studied various aspects of process capability. In one part of the study the researchers compared the variability in product of a particular piece of equipment to a known analytic capability to decide whether product consistency could be improved. The following data were obtained for 10 batches of product: (see Exercise 11.23, P. 586 for data). At the 1% significance level, do the data provide sufficient evidence to conclude that the process variation for this piece of equipment exceeds the analytic capability of 0.27? (Note: s = 0.756).

(1) H₀: ϑ = 0.27; H₁ > 0.27
(2) ∝ = 0.01
(3) χ² = 70.631
(4) Critical value: 21.666; P = 1.15 X 10 to -11 power
(5) Reject H₀
(6) At the 1% significance level, the data provide sufficient evidence to conclude that the process variation for this piece of equipment exceeds the analytical capability of 0.27.

A coffee machine is supposed to dispense 6 fluid ounces (fl oz) of coffee into a paper cup. In reality, the amounts dispensed vary from cup to cup. However, if the machine is working properly, most of the cups will contain within 10% of the advertised 6 fl oz. In other words, the standard deviation of the amounts dispensed should be less than 0.2 fl oz. A random sample of 15 cups provided the following data, in fluid ounces: (see Exercise 11.25, P. 587 for data). At the 5% significance level, do the data provide sufficient evidence to conclude that the standard deviation of the amounts being dispensed is less than 0.2 fl oz? (Note: s = 0.154).

(1) H₀: ϑ = 0.2; H₁ < 0.2
(2) ∝ = 0.05
(3) χ² = 8.317
(4) Critical value: 6.571; P = 0.128
(5) Do not reject H₀
(6) At the 5% significance level, the data do not provide sufficient evidence to conclude that the standard deviation of the amounts being dispensed is less than 0.2 fl oz.

In 2000, the mean retail price of agriculture books was $66.52. This year's retail prices for 28 randomly selected agriculture books are shown in the following table: (see Exercise 11.21, P. 586 for data). Assume that the population standard deviation of prices for this year's agriculture books is $8.45. Find a 90% confidence interval for the standard deviation of this year's retail prices of agriculture books. (Note: s = 9.229, n = 28)

$7.57 to $11.93
We can be 90% confident that the standard deviation of this year's retail prices of agriculture books is between $7.57 and $11.93.

R. Morris and E. Watson studied various aspects of process capability. In one part of the study the researchers compared the variability in product of a particular piece of equipment to a known analytic capability to decide whether product consistency could be improved. The following data were obtained for 10 batches of product: (see Exercise 11.23, P. 586 for data). Determine a 98% confidence interval for the process variation of the piece of equipment under consideration (Note: s = 0.756, n = 10).

0.49 to 1.57
We can be 98% confident that the process variation for this piece of equipment is between 0.49 and 1.57.

A coffee machine is supposed to dispense 6 fluid ounces (fl oz) of coffee into a paper cup. In reality, the amounts dispensed vary from cup to cup. However, if the machine is working properly, most of the cups will contain within 10% of the advertised 6 fl oz. In other words, the standard deviation of the amounts dispensed should be less than 0.2 fl oz. A random sample of 15 cups provided the following data, in fluid ounces: (see Exercise 11.25, P. 587 for data). Obtain a 90% confidence interval for the standard deviation of the amounts of coffee being dispensed (Note: s = 0.154, n =15).

0.119 to 0.225 fl oz
We can be 90% confident that the standard deviation of the amounts of coffee being dispensed is somewhere between 0.119 and 0.225 fl oz.

Decide whether applying one-standard-deviation χ²-procedures appears reasonable. Explain your answer.

R. Glud explores the distributions of oxygen in surface sediments from central Sagami Bay. The oxygen distribution gives important information on the general biogeochemistry of marine sediments. Measurements were performed at 16 sites. A sample of 22 depths yielded the following data, in millimoles per wquare meter per day, on diffusive oxygen uptake (DOU):
1.8, 3.3, 1.1, 2.0, 1.2, 0.7, 1.8, 3.6, 1.0, 2.3, 1.9, 1.8, 3.8, 7.6, 1.8, 3.4, 2.0, 6.7, 2.7, 1.5, 1.1, 2.0

A normal probability plot of the data suggests that the variable under consideration is far from normally distributed. So, using one-standard-deviation χ²-procedures is not reasonable.

Decide whether applying one-standard-deviation χ²-procedures appears reasonable. Explain your answer.

Trunelle examined Arizona public-company executives with salaries and bonuses totaling over $350,000. The following data provide the salaries, to the nearest thousand dollars, of a random sample of 20 such executives:

516, 770, 450, 836, 574, 680, 545, 404, 560, 672, 630, 428, 623, 745, 650, 620, 600, 450, 461, 604

A normal probability plot of the data is quite linear and reveals no outliers. So, using one-standard-deviation χ²-procedures is reasonable.

Assumptions for the one-standard-deviation χ²-test

(1) Simple random sample
(2) Normal distribution

Form of the null hypothesis for the one-standard-deviation χ²-test

H₀ : ϑ = ϑ₀

Test statistic for the one-standard deviation χ²-test

Chi-square critical value(s) calculation

df = n-1

One-standard-deviation χ²-interval procedure

df = n-1

Basic properties of F-curves

(1) the total area under an F-curve equals 1
(2) An F-curve starts at 0 on the horizontal axis and extends indefinitely to the right, approaching, but never touching, the horizontal axis as it does so.
(3) An F-curve is right skewed.

Reciprocal property of F-curves

For an F-curve with df = (v₁, v₂), the F-value having area ∝ to its left equals the reciprocal of the F-value having area ∝ to its right for an F-curve with df = (v₂, v₁).

For an F-curve with df = (60, 8), find the F-value having area 0.05 to its left.

0.24

Form of null hypothesis for two population standard deviations

H₀: ϑ₁ = ϑ₂

Test statistic for the two-standard-deviations F-test

Critical value(s) for the two-standard-deviations F-test

df = (n₁ - 1), (n₂ - 1)
** Remember reciprocal properties of F-curves (left-tail)

An F-curve has df = (12, 7). What is the number of degrees of freedom for the
(a) numerator?
(b) denominator?

(a) 12
(b) 7

An F-curve has df = (24, 30). In each case, find the F-value that has the specified area to its right.
(a) 0.05
(b) 0.01
(c) 0.025

(a) 1.89
(b) 2.47
(c) 2.14

Consider an F-curve with df = (6, 8). Obtain the F-value that has area:
(a) 0.01 to its left
(b) 0.95 to its left

(a) 0.12
(b) 3.58

Determine the two F-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas for an F-curve with
(a) df = (7, 4)
(b) df = (12, 20)

(a) 0.18, 9.07
(b) 0.33, 2.68

In using F-procedures to make inferences for two population standard deviations, why should the distributions (one for each population) of the variable under consideration be normally distributed or nearly so?

Because the procedures are based in part on the assumption that the variable under consideration is normally distributed on each population and are non-robust to violations of that assumption.

Use the two-standard-deviations F-test to conduct the required hypothesis test for comparing the standard deviations of the populations from which the samples were drawn:
s₁ = 19.4, n₁ = 41, s₂ = 10.5, n₂ = 19 ; right-tailed test, ∝ = 0.01

F = 3.41
Critical Value = 2.84
P = 0.003
Reject H₀

Use the two-standard-deviations F-test to conduct the required hypothesis test for comparing the standard deviations of the populations from which the samples were drawn:
s₁ = 28.82, n₁ = 8, s₂ = 38.97, n₂ = 13 ; left-tailed test, ∝ = 0.10

F = 0.55
Critical Value = 0.37
P = 0.216
Do not reject H₀

Use the two-standard-deviations F-test to conduct the required hypothesis test for comparing the standard deviations of the populations from which the samples were drawn:
s₁ = 14.5, n₁ = 11, s₂ = 30.4, n₂ = 9 ; two-tailed test, ∝ = 0.05

F = 0.23
Critical Values = 0.26 and 4.30
P = 0.032
Reject H₀

Algebra Exam Scores: 2 groups, one (control group) was taught the usual algebra course; the other group (experimental group) was taught by a new teaching method. The final-exam scores (out of 40 possible for the two groups are shown in Problem 11.67 P. 600.
Do the data provide sufficient evidence to conclude that there is less variation among final-exam scores when the new teaching method is used? Perform an F-test at the 5% significance level. (Note: s₁ = 7.813 and s₂ = 5.286).

H₀: ϑ₁ = ϑ₂
H₁: ϑ₁ > ϑ₂
F=2.19
∝ = 0.05
Critical value = 2.03
P = 0.035
Reject H₀
At the 5% significance level, the data provide sufficient evidence to conclude that there is less variation among final-exam scores using the new teaching method.

Videotapes of progressive relaxation exercises were shown to one group of patients and neutral videotapes to another group. Then both groups to an anxiety inventory. The data based on those results (Problem 11.69, P. 600). Do the data provide sufficient evidence to conclude that the variation in anxiety-test scores differs between patients that are shown relaxation tapes and those who are shown neutral tapes? Perform an F-test at the 10% significance level. (Note: s₁ = 10.154 and s₂ = 9.197).

H₀: ϑ₁ = ϑ₂
H₁: ϑ₁ ≠ ϑ₂
F=1.22
∝ = 0.10
Critical value = 0.53 and 1.94
P = 0.624
Do not reject H₀
At the 10% significance level, the data do not provide sufficient evidence to conclude that the variation in anxiety-test scores differs between patients seeing videotapes showing progressive relaxation exercises and those seeing neutral videotapes.

An independent golf equipment testing facility compared the difference in the performance of golf balls hit off a regular tee to those hit off a Stinger golf tee. Data on ball velocity (in mph) with each type of tee are as follows:
Stinger: x-bar₁ = 128.83, s₁ = 0.410, n₁ = 30
Regular: x-bar₂ = 127.01, s₂ = 0.894, n₂ = 30
At the 1% significance level, do the data provide sufficient evidence to conclude that the standard deviation of velocity is less with the Stinger tee than with the regular tee? (Note: For df = (29, 29), F (sub 0.01) = 2.42).

H₀: ϑ₁ = ϑ₂
H₁: ϑ₁ < ϑ₂
F=0.21
∝ = 0.01
Critical value = 0.41
P = 0.0000 (to 4 decimal places)
Reject H₀
At the 1% significance level, the data provide sufficient evidence to conclude that the standard deviation of velocity is less with the Stinger tee than with the regular tee.

What distribution is used to make inferences for one population standard deviation?

Chi-square distribution

Fill in the blanks:
(a) A χ²-curve is ________ skewed.
(b) A χ²-curve looks increasingly like a _______ curve as the number of degrees of freedom becomes larger.

(a) right
(b) normal

When you use the one-standard-deviation χ²-test or χ²-interval procedure, what assumption must be met by the variable under consideration? How important is that assumption?

The variable under consideration must be normally distributed or nearly so. It is very important because the procedures are nonrobust to violations of that assumption.

Consider a χ²-curve with 17 degrees of freedom. Use Table VII to determine:
(a) χ² (sub 0.99)
(b) χ² (sub 0.01)
(c) The χ²-value having area 0.05 to its right
(d) The χ²-value having area 0.05 to its left
(e) the two χ²-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas.

(a) 6.408
(b) 33.409
(c) 27.587
(d) 8.672
(e) 7.564, 30.191

What distribution is used to make inferences for two population standard deviations?

The F-distribution

Fill in the blanks:
(a) An F-curve is _______ skewed.
(b) For an F-curve with df = (14,5), the F-value having area 0.05 to its left equals the ________ of the F-value having area 0.05 to its right for an F-curve with df = (______, ______).
(c) The observed value of a variable having an F-distribution must be greater than or equal to ______.

(a) right
(b) reciprocal; 5, 14
(c) 0

When you use the two-standard-deviations F-test, what assumption must be met by the variable under consideration? How important is that assumption?

The distributions (one for each population) of the variable under consideration must be normally distributed or nearly so. It is very important because the procedure is nonrobust to violations of that assumption.

Consider an F-curve with df = (4, 8). Use Table VIII to determine:
(a) F (sub 0.01)
(b) F (sub 0.99)
(c) the F-value having area 0.05 to its right
(d) the F-value having area 0.05 to its left
(e) the two F-values that divide the area under the curve into a middle 0.95 area and two outside 0.025 areas.

(a) 7.01
(b) 0.07
(c) 3.84
(d) 0.17
(e) 0.11, 5.05

IQs measured on the SBS Intelligence Scale are supposed to have a standard deviation of 16 points. Twenty five randomly selected people were given the IQ test; here are the data that were obtained:
91, 102, 95, 88, 104, 96, 96, 111, 129, 118, 106, 124, 105, 112, 127, 116, 115, 101, 82, 66, 97, 121, 86, 98, 102
Assume that the IQs measured are normally distributed.
Do the data provide sufficient evidence to conclude that IQs measured on this scale have a standard deviation different from 16 points? Perform the required hypothesis test at the 10% significance level. (Note: s = 15.006).

H₀: ϑ = 16
H₁: ϑ ≠ 16
χ² = 21.110
∝ = 0.10
Critical values = 13.848 and 36.415
P = 0.736
Do not reject H₀
At the 10% significance level, the data do not provide sufficient evidence to conclude that IQs measured on this scale have a standard deviation different from 16 points.

IQs measured on the SBS Intelligence Scale are supposed to have a standard deviation of 16 points. Twenty five randomly selected people were given the IQ test; here are the data that were obtained:
91, 102, 95, 88, 104, 96, 96, 111, 129, 118, 106, 124, 105, 112, 127, 116, 115, 101, 82, 66, 97, 121, 86, 98, 102
Assume that the IQs measured are normally distributed.
Determine a 90% confidence interval for the standard deviation of IQs measured on the SBS Intelligence Scale. (Note: s = 15.006).

12.2 to 19.8 points
We can be 90% confident that the standard deviation of IQs measured on the SBS Intelligence Scale is somewhere between 12.2 and 19.8 points.

A study was conducted to decide whether elite distance runners are actually thinner than other people. The researchers measured skinfold thickness of runners and nonrunners in the same age group. The data in mm is shown in Problem 11 (P. 603-604). At the 1% significance level, do the data provide sufficient evidence to conclude that runners have less variability in skinfold thickness than others? (Note: s₁ = 1.798 and s₂ = 6.606. For df = (19, 14), F (sub 0.01) = 3.53)

H₀: ϑ₁ = ϑ₂
H₁: ϑ₁ < ϑ₂
F = 0.07
∝ = 0.01
Critical value = 0.28
P = 0.00000579
Reject H₀
At the 1% significance level, the data provide sufficient evidence to conclude that runners have less variability in skinfold thickness than others.

population proportion

The proportion (percentage) of the entire population that has the specified attribute (denoted p)

sample proportion

The proportion (percentage) of a sample from the population that has the specified attribute (denoted p-hat).

formula for computing sample proportion

One-proportion z-interval procedure

Margin of error for the estimate of p

Sample size for estimating p if you cannot make an educated guess

Rounded up to whole number!

Sample size for estimating p if you can make an educated guess

with an educated guess for value of p-hat, rounded up to whole number

Since 1966, 45% of the No. 1 draft picks in the NBA have been centers.
(a) identify the population
(b) identify the specified attribute
(c) is the proportion 0.45 (45%) a population proportion or a sample proportion? Explain your answer.

(a) the No. 1 draft picks in the NBA since 1966
(b) being a center
(c) population proportion. It is the proportion of the population of No. 1 draft picks in the NBA since 1996 who are centers.

A Harris Poll asked Americans whether states should be allowed to conduct random drug tests on elected officials. Of 21,355 respondents, 79% said yes.
(a) Determine the margin of error for a 99% confidence interval
(b) Without doing any calculations, indicate whether the margin of error is larger or smaller for a 90% confidence interval. Explain your answer.

(a) 0.00718
(b) smaller. The higher the confidence level, the higher the margin of error and the wider the confidence interval.

In each of parts (a)-(c), we have given a likely range for the observed value of a sample proportion p-hat. Based on the given range, identify the educated guess that should be used for the observed value of p-hat to calculate the required sample size for a prescribed confidence level and margin of error.
(a) 0.2 to 0.4
(b) 0.2 or less
(c) 0.4 or greater
(d) In each of parts (a)-(c), which observed values of the sample proportion will yield a larger margin of error than the one specified if the educated guess is used for the sample size computation?

(a) 0.4
(b) 0.2
(c) 0.5
(d) 0.4 < p-hat < 0.6 (a)
0.2 < p-hat < 0.8 (b)
none (c)

Given the number of successes and the sample size for a simple random sample from a population:
x = 8, n = 40, 95% level
(a) Determine the sample proportion
(b) Decide whether using the one-proportion z-interval procedure is appropriate
(c) If appropriate, use the one-proportion z-interval procedure to find the confidence interval at the specified confidence level.

(a) p-hat = 0.2
(b) appropriate
(c) 0.076 to 0.324

Given the number of successes and the sample size for a simple random sample from a population:
x = 35, n = 50, 99% level
(a) Determine the sample proportion
(b) Decide whether using the one-proportion z-interval procedure is appropriate
(c) If appropriate, use the one-proportion z-interval procedure to find the confidence interval at the specified confidence level.

(a) p-hat = 0.7
(b) appropriate
(c) 0.533 to 0.867

Given the number of successes and the sample size for a simple random sample from a population:
x = 16, n = 20, 90% level
(a) Determine the sample proportion
(b) Decide whether using the one-proportion z-interval procedure is appropriate
(c) If appropriate, use the one-proportion z-interval procedure to find the confidence interval at the specified confidence level.

(a) p-hat = 0.8
(b) not appropriate

A total of 603 interviews were conducted among a national sample of adults with household incomes of at least $150,000. Of the adults interviewed, 410 said they had make a purchase online in the past year. Find a 95% confidence interval for the proportion of all U.S. adults with household incomes of at least $150,000 who made an online purchase in the past year.

0.643 to 0.717
We can be 95% confident that the proportion of all U.S. adults with household incomes of at least $150,000 who made an online purchase in the past year is somewhere between 0.643 and 0.717.

Studies are performed to estimate the percentage of the nation's 10 million asthmatics who are allergic to sulfites. In one survey, 38 of 500 randomly selected U.S. asthmatics were found to be allergic to sulfites. Find a 95% confidence interval for the proportion, p, of all U.S. asthmatics who are allergic to sulfites. Interpret your result.

0.0528 to 0.0992
We can be 95% confident that the proportion of all U.S. asthmatics who are allergic to sulfites is somewhere between 0.0528 and 0.0992.

In a recent survey of 1000 registered voters, 80% favored the creation of standards to limit such pollution by CAFOs and in general viewed CAFOs unfavorably. Find a 99% confidence interval for the percentage of all registered voters who favor the creation of standards on CAFO pollution and, in general, view CAFOs unfavorably. Interpret your result.

76.7% to 83.3%
We can be 99% confident that the percentage of all registered voters who favor the creation of standards on CAFO pollution and, in general, view CAFOs unfavorably is somewhere between 76.7% and 83.3%.

An ABC News poll found that Americans now have relatively warm feelings toward Russia, a former adversary. The poll, conducted by telephone among a random sample of 1043 adults, found that 647 of those sampled consider the two countries friends. The margin of error for the poll was plus or minus 2.9 percentage points (for a 0.95 confidence level). Use this information to obtain a 95% confidence interval for the percentage of all Americans who consider the two countries friends.

59.1% to 64.9%

A Harris Poll taken in 2005 of 1010 U.S. adults found that 444 of them approved of the way that President George W. Bush was doing his job. Find and interpret a 95% confidence interval for the proportion of all U.S. adults who, at the time, approved of President Bush.

0.409 to 0.470. We can be 95% confident that the proportion of all U.S. adults who, at the time, approved of President Bush was somewhere between 0.409 and 0.470.

In a recent survey of 1000 registered voters, 80% favored the creation of standards to limit such pollution by CAFOs and in general viewed CAFOs unfavorably.
(a) Determine the margin of error for the estimate of the percentage.
(b) Obtain a sample size that will ensure a margin of error of at most 1.5 percentage points for a 99% confidence interval without making a guess for the observed value of p-hat.

(a) 3.3% (i.e., 3.3 percentage points)
(b) 7368

Studies are performed to estimate the percentage of the nation's 10 million asthmatics who are allergic to sulfites. In one survey, 38 of 500 randomly selected U.S. asthmatics were found to be allergic to sulfites.
(a) Determine the margin of error for the estimate of p.
(b) Obtain a sample size that will ensure a margin of error of at most 0.01 for a 95% confidence interval without making a guess for the observed value of p-hat.

(a) 0.0232
(b) 9604

Assumptions for the one-proportion z-test

(1) Simple random sample
(2) Both np₀ and n(1-p₀) are 5 or greater

Format of the null hypothesis for the one-proportion z-test

H₀: p = p₀

Test statistic for the one-proportion z-test

Critical value(s) for one-proportion z-test

Given the number of successes and the sample size for a simple random sample from a population:
x = 8, n = 40, H₀: p = 0.3, H₁: p < 0.3, ∝ = 0.10
(a) Determine the sample proportion
(b) Decide whether using the one-proportion z-test is appropriate
(c) If appropriate, use the one-proportion z-test to perform the specified hypothesis test.

(a) p-hat = 0.2
(b) appropriate
(c) z = -1.38; critical value = -1.28; p-value = 0.084
Reject H₀

Given the number of successes and the sample size for a simple random sample from a population:
x = 35, n = 50, H₀: p = 0.6, H₁: p > 0.6, ∝ = 0.05
(a) Determine the sample proportion
(b) Decide whether using the one-proportion z-test is appropriate
(c) If appropriate, use the one-proportion z-test t

(a) p-hat = 0.7
(b) appropriate
(c) z = 1.44, critical value = 1.645; p-value = 0.074
Do not reject H₀

Given the number of successes and the sample size for a simple random sample from a population:
x = 16, n = 20, H₀: p = 0.7, H₁: p ≠ 0.7, ∝ = 0.05
(a) Determine the sample proportion
(b) Decide whether using the one-proportion z-test is appropriate
(c) If appropriate, use the one-proportion z-test t

(a) p-hat = 0.8
(b) appropriate
(c) z = 0.98; critical values = ±1.96; p-value = 0.329
Do not reject H₀

People who were born between 1978 and 1983 are sometimes classified by demographers as belonging to Generation Y. According to a recent survey of 850 Generation Y web users, 459 reported using the Internet to download music.
(a) Determine the sample proportion.
(b) At the 5% significance level, do the data provide sufficient evidence to conclude that a majority of Generation Y web users use the Internet to download music?

(a) p-hat = 0.54
(b) H₀: p = 0.5; H₁: p > 0.5
∝ = 0.05
z = 2.33
critical value = 1.645
p-value = 0.0099
Reject H₀
At the 5% significance level, the data provide sufficient evidence to conclude that a majority of Generation Y web users use the Internet to download music.

According to a publication, 13.6% of 18-25-year-olds were current users of marijuana or hashish in 2000. A recent poll of 1283 randomly selected 18-25-year-olds revealed that 205 currently use marijuana or hashish. At the 10% significance level, do the data provide sufficient evidence to conclude that the percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 2000 percentage of 13.6%?

H₀: p = 0.136; H₁: p ≠ 0.136
∝ = 0.10
z = 2.49
critical values = ±1.645
p-value = 0.0128
Reject H₀
At the 10% significance level, the data provide sufficient evidence to conclude that the percentage of 18-25-year-olds who currently use marijuana or hashish has changed from the 2000 percentage of 13.6%.

A CNN poll conducted in September 2005 had the headline "Most Americans Believe New Orleans Will Never Recover". Of 609 adults polled by telephone, 341 said they believe the hurricane devastated the city beyond repair. At the 1% significance level, do the data provide sufficient evidence to justify the headline? Explain your answer

H₀: p = 0.5; H₁: p > 0.5
∝ = 0.01
z = 2.96
critical value = 2.33
p-value = 0.002
Reject H₀
At the 1% significance level, the data provide sufficient evidence to conclude that most Americans believe New Orleans will never recover.

L. Bugeja and R. Franklin examined drowning deaths of young children in Victorian dams. Of 11 young children who drowned in Victorian dams located on farms, 5 were girls. At the 5% significance level, do the data provide sufficient evidence to conclude that, of all young children drowning in Victorian dams located on farms, less than half are girls?

H₀: p = 0.5; H₁: p < 0.5
∝ = 0.05
z = -0.30
critical value = -1.645
p-value = 0.382
Do not reject H₀
At the 5% significance level, the data do not provide sufficient evidence to conclude that, of all young children drowning in Victorian dams located on farms, less than half are girls.

Assumptions for two-proportions z-test

(1) Simple random samples
(2) Independent samples
(3) x₁, n₁-x₁, x₂, and n₂-x₂ are all 5 or greater

Form of null hypothesis for two-proportions z-test

H₀: p₁ = p₂

Calculation formula for test statistic for two-proportions z-test

Pp = (x₁ + x₂) / (n₁ + n₂)

Critical value calculation for two-proportions z-test

Two-proportions z-interval procedure

Margin of error for the estimate of p₁-p₂

Industry Research polled teenagers on sunscreen use. The survey revealed that 46^ of teenage girls and 30% of teenage boys regularly use sunscreen before going out in the sun.
(a) Identify the specified attribute.
(b) Identify the two populations
(c) Are the proportions 0.46 (46%) and 0.30 (30%) sample proportions or population proportions? Explain your answer.

(a) uses sunscreen before going out in the sun
(b) teenage girls and teenage boys
(c) sample proportions. Industry Research acquired those proportions by polling samples of the populations of all teenage girls and all teenage boys.

Given the number of successes and the sample sizes for independent simple random samples from two populations:
x₁=18, n₁=40, x₂=30, n₂=40; left-tailed test, ∝=0.10; 80% confidence interval
(a) Determine the sample proportions
(b) Decide whether using the two-proportions z-procedures is appropriate. If so, also do parts (c) and (d).
(c) Use the two-proportions z-test to conduct the required hypothesis test
(d) Use the two-proportions z-interval procedure to find the specified confidence interval.

(a) p-hat₁ = 0.45, p-hat₂ = 0.75
(b) appropriate
(c) z = -2.74; critical value = -1.28; p-value = 0.003; Reject H₀
(d) -0.434 to -0.166

Given the number of successes and the sample sizes for independent simple random samples from two populations:
x₁=15, n₁=20, x₂=18, n₂=30; right-tailed test, ∝=0.05; 90% confidence interval
(a) Determine the sample proportions
(b) Decide whether using the two-proportions z-procedures is appropriate. If so, also do parts (c) and (d).
(c) Use the two-proportions z-test to conduct the required hypothesis test
(d) Use the two-proportions z-interval procedure to find the specified confidence interval.

(a) p-hat₁ = 0.75, p-hat₂ = 0.60
(b) appropriate
(c) z = 1.10; critical value = 1.645; p-value = 0.136; do not reject H₀
(d) -0.067 to 0.367

Given the number of successes and the sample sizes for independent simple random samples from two populations:
x₁=30, n₁=80, x₂=15, n₂=20; two-tailed test, ∝=0.05; 95% confidence interval
(a) Determine the sample proportions
(b) Decide whether using the two-proportions z-procedures is appropriate. If so, also do parts (c) and (d).
(c) Use the two-proportions z-test to conduct the required hypothesis test
(d) Use the two-proportions z-interval procedure to find the specified confidence interval.

(a) p-hat₁ = 0.375, p-hat₂ = 0.750
(b) appropriate
(c) z = -3.02; critical values = ±1.96; p-value = 0.003; Reject H₀
(d) -0.592 to -0.158

For a study, doctors enrolled women prior to conception and divided them randomly into two groups. One group, consisting of 2701 women, took daily multivitamins containing 0.8 mg of folic acid; the other group, consisting of 2052 women, received only trace elements. Major birth defects occurred in 35 cases when the women took folic acid and in 47 cases when the women did not.
(a) At the 1% significance level, do the data provide sufficient evidence to conclude that women who take folic acid are at lesser risk of having children with major birth defects?

H₀: p₁ = p₂; H₁: p₁ < p₂
∝ = 0.01
z = -2.61
critical value = -2.33
p = 0.0045
Reject H₀
At the 1% significance level, the data provide sufficient evidence to conclude that women who take folic acid are at lesser risk of having children with major birth defects.

Response Insurance collects data on seat-belt use among U.S. drivers. Of 1000 drivers 25-34 years old, 27% said that they buckle up, whereas 330 of 1100 drivers 45-64 years old said that they did. At the 10% significance level, do the data suggest that there is a difference in seat-belt use between drivers 25-34 years old and those 45-64 years old?

H₀: p₁ = p₂; H₁: p₁ ≠ p₂
∝ = 0.10
z = -1.52
critical values = ± 1.645
p = 0.1286
Do not reject H₀
At the 10% significance level, the data do not provide sufficient evidence to conclude that there is a difference in seat-belt use between drivers who are 25-34 years old and drivers who are 45-64 years old.

According to the HHS, for adults, a BMI of greater than 25 indicates an above healthy weight. Of 750 randomly selected adults whose highest degree is a bachelors, 386 have an above healthy weight; and of 500 randomly selected adults with a graduate degree, 237 have an above healthy weight. Determine at the 5% significance level, whether the percentage of adults who have an above healthy weight is greater for those whose highest degree is a bachelors than for those with a graduate degree.

H₀: p₁ = p₂; H₁: p₁ > p₂
∝ = 0.05
z = 1.41
critical value = 1.645
p = 0.0793
Do not reject H₀
At the 5% significance level, the data do not provide sufficient evidence to conclude that the percentage who are overweight is greater for those whose highest degree is a bachelors than for those with a graduate degree.

For a study, doctors enrolled women prior to conception and divided them randomly into two groups. One group, consisting of 2701 women, took daily multivitamins containing 0.8 mg of folic acid; the other group, consisting of 2052 women, received only trace elements. Major birth defects occurred in 35 cases when the women took folic acid and in 47 cases when the women did not.
Determine and interpret a 98% confidence interval for the difference between the rates of major birth defects for babies born to women who have taken folic acid and those born to women who have not.

-0.0191 to -0.000746
We can be 98% confident that the rate of major birth defects for babies born to women who have not taken folic acid is somewhere between -0.0191 and -0.000746 (or 1 per thousand and 19 per thousand).

Response Insurance collects data on seat-belt use among U.S. drivers. Of 1000 drivers 25-34 years old, 27% said that they buckle up, whereas 330 of 1100 drivers 45-64 years old said that they did. Find and interpret a 90% confidence interval for the difference between the proportions of seat-belt users for drivers in the age groups 25-34 years and 45-64 years.

-0.0624 to 0.00240
We can be 90% confident that the difference between the proportions of seat-belt users for drivers in the age groups 25-34 years and 45-64 years is somewhere between -0.0624 and 0.00240.

According to the HHS, for adults, a BMI of greater than 25 indicates an above healthy weight. Of 750 randomly selected adults whose highest degree is a bachelors, 386 have an above healthy weight; and of 500 randomly selected adults with a graduate degree, 237 have an above healthy weight. Determine and interpret a 90% confidence interval for the difference between the percentages of adults in the two degree categories who have an above healthy weight.

-0.68% to 8.81%

Independent random samples were taken of 300 U.S. women and 250 Canadian women. Of the U.S. women, 219 were found to be in the labor force; and of the Canadian women, 174 were found to be in the labor force. At the 5% significance level, do the data suggest that there is a difference between the labor-force participation rates of U.S. and Canadian women?

H₀: p₁ = p₂; H₁: p₁ ≠ p₂
∝ = 0.05
z = 0.88
critical values = ±1.96
p = 0.379
Do not reject H₀
At the 5% significance level, the data do not provide sufficient evidence to conclude that there is a difference between the labor-force participation rates of U.S. and Canadian women.

Independent random samples were taken of 300 U.S. women and 250 Canadian women. Of the U.S. women, 219 were found to be in the labor force; and of the Canadian women, 174 were found to be in the labor force. Find and interpret a 95% confidence interval for the difference between the labor-force participation rates of U.S. and Canadian women.

-0.042 to 0.110
We can be 95% confident that the difference between the labor-force participation rates of U.S. and Canadian women is somewhere between -4.2% and 11.0%.

A Harris poll was conducted to estimate the proportion of Americans who feel that marijuana should be legalized for medicinal use in patients with cancer and other painful and terminal diseases. Identify the:
(a) specified attribute
(b) population
(c) population proportion
(d) According to the poll, 80% of the 83,957 respondents said that marijuana should be legalized for medicinal use. Is the proportion 0.80 (80%) a sample proportion or a population proportion? Explain your answer.

(a) feeling that marijuana should be legalized for medicinal use in patients with cancer and other painful and terminal diseases
(b) Americans
(c) Proportion of all Americans who feel that marijuana should be legalized for medicinal use in patients with cancer and other painful and terminal diseases
(d) sample proportion. It is the proportion of Americans sampled who feel that marijuana should be legalized for medicinal use in patients with cancer and other painful and terminal diseases.

Why is a sample proportion generally used to estimate a population proportion instead of obtaining the population proportion directly?

Generally, obtaining a sample proportion can be done more quickly and is less costly than obtaining the population proportion. Sampling is often the only practical way to proceed.

Explain what each phrase means in the context of inferences for a population proportion:
(a) number of successes
(b) number of failures

(a) the number of members in the sample that have the specified attribute
(b) the number of members in the sample that do not have the specified attribute

Fill in the blanks:
(a) The mean of all possible sample proportions is equal to the ________.
(b) For large samples, the possible sample proportions have approximately a ________ distribution.
(c) A rule of thumb for using a normal distribution to approximate the distribution of all possible sample proportions is that both ______ and ______ are ________ or greater.

(a) population proportion
(b) normal
(c) np, n(1-p), 5

What does the margin of error for the estimate of a population proportion tell you?

The precision with which a sample proportion, p-hat, estimates the population proportion, p, at the specified confidence level.

A poll was conducted by Opinion Research Corporation to estimate the proportions of men and women who get the "holiday blues". Identify the:
(a) specified attribute
(b) two populations
(c) two population proportions
(d) two sample proportions
(e) According to the poll, 34% of men and 44% of women get the "holiday blues". Are the proportions 0.34 and 0.44 sample proportions or population proportions? Explain your answer.

(a) Getting the "holiday blues"
(b) all men, all women
(c) The proportion of all men who get the "holiday blues" and the proportion of all women who get the "holiday blues"
(d) The proportion of all sampled men who get the "holiday blues" and the proportion of all sampled women who get the "holiday blues"
(e) Sample proportions. The poll used samples of men and women to obtain the proportions.

Suppose that you are using independent samples to compare two population proportions. Fill in the blanks:
(a) The mean of all possible differences between the two sample proportions equals the __________.
(b) For large samples, the possible differences between the two sample proportions have approximately a ________ distribution.

(a) difference between the population proportions
(b) normal

In 2002, a poll asked U.S> adults whether they would get a smallpox shot if it were available. When the risk of the vaccine was described in detail, 4 in 10 of those surveyed said they would take the smallpox shot. According to the article, "the results have a three-point margin of error" (for a 0.95 confidence level). Use the information provided to obtain a 95% confidence interval for the percentage of all U.S. adults who would take a smallpox shot, knowing the risk of the vaccine.

37.0% to 43.0%

Suppose that you want to find a 95% confidence interval based on independent samples for the difference between two population proportions and that you want a margin of error of at most 0.01.
(a) Without making an educated guess for the observed sample proportions, find the required common sample size.
(b) Suppose that, from past experience, you are quite sure that the two sample proportions will be 0.75 or greater. What common sample size should you use?

(a) 19,208
(b) 14,406

The National Association of Colleges and Employers sponsors the Graduating Student and Alumni Survey. Part of the survey gauges student optimism in landing a job after graduation. According to one year's survey results, among the 1218 respondents, 733 said that they expected difficulty finding a job. Use these data to obtain and interpret a 95% confidence interval for the proportion of students who expect difficulty finding a job.

0.574 to 0.629. We can be 95% confident that the percentage of students who expect difficulty finding a job is somewhere between 57.4% and 62.9%.

In an issue of Parade Magazine, the editors reported on a national survey on law and order. One question asked of the 2512 U.S. adults who took part was whether they believed that juries "almost always" convict the guilty and free the innocent. Only 578 said that they did. Do the data provide sufficient evidence to conclude that less than one in four Americans believe that juries "almost always" convict the guilty and free the innocent? Perform the appropriate hypothesis test at the 5% significance level. Assess the strength of the evidence against the null hypothesis.

H₀: p = 0.25, H₁: p < 0.25
∝ = 0.05
z = -2.30
critical value = -1.645
p = 0.0107
Reject H₀
At the 5% significance level, the data provide sufficient evidence to conclude that less than one in four Americans believe that juries "almost always" convict the guilty and free the innocent. The data provide strong evidence against the null hypothesis.

The NCAA Men's Division I Basketball Championship is held each spring and features 65 college basketball teams. This 20-day tournament is colloquially known as "March Madness." A Harris Poll, conducted in March, 2006, asked 2435 randomly selected U.S. adults whether they would participate in an office pool for March Madness; 317 said they would. Use these data to find and interpret a 95% confidence interval for the percentage of U.S. adults who would participate in an office pool for March Madness.

11.7% to 14.4%
We can be 95% confident that the percentage of U.S. adults who would participate in an office pool for March Madness is somewhere between 11.7% and 14.4%.

In a January, 2006 Harris Poll of 1961 randomly selected U.S. adults, 1137 said that they do not believe that abstinence programs are effective in reducing or preventing AIDS. At the 5% significance level, do the data provide sufficient evidence to conclude that a majority of all U.S. adults feel that way?

H₀: p = 0.5, H₁: p > 0.5
∝ = 0.05
z = 7.07
critical value = 1.645
p = 0.0000 (to 4 decimal places)
Reject H₀
At the 5% significance level, the data provide sufficient evidence to conclude that a majority of U.S. adults do not believe that abstinence programs are effective in reducing or preventing AIDS.

N. Hill conducted a clinical study to compare the standard treatment for head lice infestation with the Bug Buster kit. For the study, 56 patients were randomly assigned to use the Bug Buster kit and 70 were assigned to use the standard treatment. Thirty-two patients in the Bug Buster kit group were cured, whereas nine of those in the standard treatment group were cured.
(a) At the 5% significance level, do these data provide sufficient evidence to conclude that a difference exists in the cure rates of the two types of treatment?
(b) Determine a 95% confidence interval for the difference in cure rates for the two types of treatment.

H₀: p₁ = p₂, H₁: p₁ ≠ p₂
∝ = 0.05
z = 5.27
critical values = ±1.96
p = 0.0000 (to 4 decimal places)
Reject H₀
At the 5% significance level, the data provide sufficient evidence to conclude that a difference exists in the cure rates of the two types of treatment.

(b) 0.291 to 0.594. We can be 95% confident that use of the Bug Buster kit will increase the proportion of those cured by somewhere between 0.291 and 0.594.

chi-square goodness-of-fit test

the procedure used to perform a hypothesis test about the distribution of a qualitative (categorical) variable or a discrete quantitative variable that has only finitely many possible values. It can be applied to any variable whose possible values have been grouped into a finite number of categories.

Assumptions for the chi-square goodness-of-fit test

(1) All expected frequencies are 1 or greater
(2) At most 20% of the expected frequencies are less than 5
(3) Simple random sample

Forms of the null and alternative hypotheses for the chi-square goodness-of-fit test

H₀: The variable has the specified distribution
H₁: The variable does not have the specified distribution

How to calculate expected frequencies for the chi-square goodness-of-fit test

E = np
where n = sample size
and p = relative frequency (or probability) given for the value in the null hypothesis

How to calculate the test statistic for the chi-square goodness-of-fit test

How to calculate the critical value for the chi-square goodness-of-fit test

For a chi-square goodness-of-fit test, what is meant by the phrase "goodness-of-fit"?

The hypothesis test is carried out by determining how well the observed frequencies fit the expected frequencies.

Given the relative frequencies for the null hypothesis of a chi-square goodness-of-fit test and the sample size, decide whether Assumptions 1 and 2 for using that test are satisfied.

Sample size: n = 100
Relative frequencies: 0.65, 0.30, 0.05

Both assumptions are satisfied.

Given the relative frequencies for the null hypothesis of a chi-square goodness-of-fit test and the sample size, decide whether Assumptions 1 and 2 for using that test are satisfied.

Sample size: n = 50
Relative frequencies: 0.20, 0.20, 0.25, 0.30, 0.05

Both assumptions are satisfied. Note that 20% of the expected frequencies are less than 5.

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